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http://mathhelpboards.com/math-notes-49/inverse-sine-tangent-integrals-related-functions-6483.html

- Thread starter MarkFL
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http://mathhelpboards.com/math-notes-49/inverse-sine-tangent-integrals-related-functions-6483.html

- Jan 17, 2013

- 1,667

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

- Sep 16, 2013

- 337

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

Damn!!! And there was I just about to go to bed and all... You a bad mammal!

More seriously though, this looks an interesting problem. I'll get "on it like a car bonnet" tomorrow, and then - internet connection allowing - post a reply.

Cheers Z!!

- - - Updated - - -

Incidentally, it looks like a simple differentiation will get me most of the way there, since

\(\displaystyle \frac{d}{dz}\text{Ti}_{m+1}(z)=\frac{\text{Ti}_m(z)}{z}\)

- Sep 16, 2013

- 337

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

Here's my solution...

To start with, make the substitution \(\displaystyle y=at\,\) to obtain

\(\displaystyle \frac{1}{a}\int_0^a\frac{(1-y^2)}{(1+y^2)^2}\,\log^2(y/a)\,dy=\)

\(\displaystyle \frac{1}{a}\int_0^a\frac{2-(1+y^2)}{(1+y^2)^2}\,\log^2(y/a)\,dy=\)

\(\displaystyle \frac{2}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)^2}\,dy-

\frac{1}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy=\)

\(\displaystyle \frac{2}{a}\int_0^a\frac{[(1+y^2)-y^2]\log^2(y/a)}{(1+y^2)^2}\,dy-

\frac{1}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy=\)

\(\displaystyle \left(\frac{2}{a}-\frac{1}{a}\right)\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-\frac{2}{a}\int_0^a\frac{y^2\log^2(y/a)}{(1+y^2)^2}\,dy=\)

\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-\frac{2}{a}\int_0^a\frac{y^2\log^2(y/a)}{(1+y^2)^2}\,dy=\)

\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy+\frac{1}{a}\int_0^a\left[\frac{-2y}{(1+y^2)^2}\right]y\,\log^2(y/a)\,dy\)

The term in large square brackets in the last integral is the derivative:

\(\displaystyle \frac{d}{dy}\frac{1}{(1+y^2)}=\frac{-2y}{(1+y^2)^2}\)

So we can perform an integration by parts on that last integral:

\(\displaystyle \int_0^a\left[\frac{-2y}{(1+y^2)^2}\right]y\,\log^2(y/a)\,dy=\)

\(\displaystyle \frac{y\,\log^2(y/a)}{(1+y^2)}\,\Biggr|_0^a-\int_0^a\frac{1}{(1+y^2)}\left[\log^2(y/a)+2\log(y/a)\right]\,dy=\)

\(\displaystyle -\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-2\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy\)

Inserting this back into the previous partial evaluation we get

\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy+\frac{1}{a}\left[-\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-2\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy\right]=\)

\(\displaystyle -\frac{2}{a}\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy=\)

\(\displaystyle -\frac{2}{a}\,\left[\tan^{-1}y\log(y/a)\,\Biggr|_0^a-\frac{1}{a}\,\int_0^a\frac{\tan^{-1}y}{(y/a)}\,dy\right]=\)

\(\displaystyle \frac{2}{a}\int_0^a\frac{\tan^{-1}y}{y}\,dy=\frac{2}{a}\,\text{Ti}_2(a)\, .\, \Box\)

That was really good fun, Z! If you have any others, I'll happily give them a go... Not saying I'll get them, but always up for a go...

Cheers!!